Course: 11.5 遗传序列和从共同差异和任期中寻找n期

Section outline

Select section General

Collapse Expand
General

Collapse all Expand all
- Select activity 新闻通告
  
  新闻通告 Forum
Select section 具有共同差异和任期的理论性序列和查找第n期

Collapse Expand
具有共同差异和任期的理论性序列和查找第n期
Halley's Comet appears in the sky approximately every 76 years. The comet was first spotted in the year 1531. Find the $\begin{align*}n^{th}\end{align*}$ term rule and the $\begin{align*}10^{th}\end{align*}$ term for the represented by this situation.
::Halley的彗星大约每隔76年在天空中出现一次。彗星是1531年首次发现的。找到以这种情形为代表的星号的第n期规则和第10期规则。

Arithmetic Sequence
::亚学序列

In this concept we will begin looking at a specific type of sequence called an arithmetic sequence . In an arithmetic sequence the difference between any two consecutive terms is constant. This constant difference is called the common difference . We can generalize the equation for an arithmetic sequence below:
::在此概念中, 我们将开始查看一种特定类型的序列, 称为算术序列。在算术序列中, 任何两个连续任期之间的差是不变的。这个不变的差称为共同差。我们可以在下面的算术序列中将方程式概括为 :

$\begin{align*}a_n-a_{n-1}=d\end{align*}$ , where $\begin{align*}a_{n-1}\end{align*}$ and $\begin{align*}a_n\end{align*}$ represent two consecutive terms and $\begin{align*}d\end{align*}$ represents the common difference.
::a-an-1=d,其中an-1代表连续两届,d代表共同差额。

Since the same value, the common difference, $\begin{align*}d\end{align*}$ , is added to get each successive term in an arithmetic sequence we can determine the value of any term from the first term and how many time we need to add $\begin{align*}d\end{align*}$ to get to the desired term as illustrated below:
::由于同值,加上了共同差数(d),以便每个连续学期都按算术顺序计算,我们可以确定从第一个学期起任何学期的价值,以及我们需要多少时间来增加 d ,以便达到下列预期学期:

Given the sequence: $\begin{align*}22, 19, 16, 13, \ldots\end{align*}$ in which $\begin{align*}a_1=22\end{align*}$ and $\begin{align*}d=-3\end{align*}$
::根据顺序: 22,19,16,13... 其中a1=22和d3

$\begin{aligned} a_{1} & = 22 o r 22 + (1 - 1) (- 3) = 22 + 0 = 22 \\ a_{2} & = 19 o r 22 + (2 - 1) (- 3) = 22 + (- 3) = 19 \\ a_{3} & = 16 o r 22 + (3 - 1) (- 3) = 22 + (- 6) = 16 \\ a_{4} & = 13 o r 22 + (4 - 1) (- 3) = 22 + (- 9) = 13 \\ ⋮ \\ a_{n} & = 22 + (n - 1) (- 3) \\ a_{n} & = 22 - 3 n + 3 \\ a_{n} & = - 3 n + 25 \end{aligned}$
$\begin{align*}a_1&=22 \ or \ 22+(1-1)(-3)=22+0=22 \\ a_2&=19 \ or \ 22+(2-1)(-3)=22+(-3)=19 \\ a_3&=16 \ or \ 22+(3-1)(-3)=22+(-6)=16 \\ a_4&=13 \ or \ 22+(4-1)(-3)=22+(-9)=13 \\ &\qquad \qquad \vdots \\ a_n&=22+(n-1)(-3) \\ a_n&=22-3n+3 \\ a_n&=-3n+25 \end{align*}$
::a1=22或22+(1-1)(-3)=22+0=22a2=19或22+(2-1)(3)=22+(3)=22+(3)=19a3=16或22+(3-1)(3)=22+(6)=16a4=13或22+(4-1)(3)=22+(9)(9)=13an=22+(n-1)(3)an=22-3n+3n+3an_3n+3n+3n+25

Now we can generalize this into a rule for the $\begin{align*}n^{th}\end{align*}$ term of any arithmetic sequence:
::现在我们可以把它概括为任何算术序列的第 n 术语的规则:

$\begin{aligned} a_{n} = a_{1} + (n - 1) d \end{aligned}$
$\begin{align*}a_{n}=a_1+(n-1)d\end{align*}$
::a=a1+(n-1)d

Let's find the common difference and $\begin{align*}n^{th}\end{align*}$ term rule for the arithmetic sequence: $\begin{align*}2, 5, 8, 11 \ldots\end{align*}$
::让我们找出算术序列的共同的差别和nth术语规则: 2,5,8,11...

To find the common difference we subtract consecutive terms.
::为了找到共同的差别,我们连续减一减。

$\begin{aligned} 5 - 2 & = 3 \\ 8 - 5 & = 3, thus the common difference is 3. \\ 11 - 8 & = 3 \end{aligned}$
$\begin{align*}5-2&=3 \\ 8-5&=3 \ ,\text{thus the common difference is} \ 3. \\ 11-8&=3\end{align*}$
::5-2=38-5=3,因此共同差为3.11-8=3

Now we can put our first term and common difference into the $\begin{align*}n^{th}\end{align*}$ term rule discovered above and simplify the expression.
::现在我们可以把我们的第一个任期和共同的区别放在上面发现的第n个任期规则中并简化表达方式

$\begin{aligned} a_{n} & = 2 + (n - 1) (3) \\ = 2 + 3 n - 3, so a_{n} = 3 n - 1. \\ = 3 n - 1 \end{aligned}$
$\begin{align*}a_n&=2+(n-1)(3) \\ &=2+3n-3 \quad ,\text{so} \ a_n=3n-1. \\ &=3n-1\end{align*}$
::an=2+(n-1)(3)=2+3n-3,所以 an=3n-1.=3n-1

Now, let's find the $\begin{align*}n^{th}\end{align*}$ term rule and thus the $\begin{align*}100^{th}\end{align*}$ term for the arithmetic sequence in which $\begin{align*}a_1=-9\end{align*}$ and $\begin{align*}d=2\end{align*}$ .
::现在,让我们来看看 nth 术语规则, 以及计算序列的第 100 个术语, 其中 a1\\\ 9 和 d= 2 。

We have what we need to plug into the rule:
::我们有我们需要的插入规则:

$\begin{aligned} a_{n} & = - 9 + (n - 1) (2) \\ = - 9 + 2 n - 2, thus the n^{t h} term rule is a_{n} = 2 n - 11. \\ = 2 n - 11 \end{aligned}$
$\begin{align*}a_n&=-9+(n-1)(2) \\ &=-9+2n-2 \quad , \text{thus the} \ n^{th} \ \text{term rule is} \ a_n=2n-11. \\ &=2n-11\end{align*}$
::an9+(n- 1)(2)9+2n-2n-2,因此 nth 术语规则是 an=2n- 11.=2n- 11

Now to find the $\begin{align*}100^{th}\end{align*}$ term we can use our rule and replace $\begin{align*}n\end{align*}$ with 100: $\begin{align*}a_{100}=2(100)-11=200-11=189\end{align*}$ .
::现在找到第100个术语,我们可以使用我们的规则,用100:a=2(100)-11=200-11=189取代n。

Finally, let's find the $\begin{align*}n^{th}\end{align*}$ term rule and thus the $\begin{align*}100^{th}\end{align*}$ term for the arithmetic sequence in which $\begin{align*}a_3=8\end{align*}$ and $\begin{align*}d=7\end{align*}$ .
::最后,让我们找到 nth 术语规则, 以及计算序列的第100个术语, 其中 a3=8 和 d=7 。

This one is a little less straightforward as we will have to first determine the first term from the term we are given. To do this, we will replace $\begin{align*}a_n\end{align*}$ with $\begin{align*}a_3=8\end{align*}$ and use 3 for $\begin{align*}n\end{align*}$ in the formula to determine the unknown first term as shown:
::这一点不太直截了当,因为我们首先必须确定从我们被赋予的这一任期开始的第一个任期。为此,我们将用A3=8取代A3=8,并在公式中用n3代替n,以确定未知的第一个任期,如下所示:

$\begin{aligned} a_{1} + (3 - 1) (7) & = 8 \\ a_{1} + 2 (7) & = 8 \\ a_{1} + 14 & = 8 \\ a_{1} & = - 6 \end{aligned}$
$\begin{align*}a_1+(3-1)(7)&=8 \\ a_1+2(7)&=8 \\ a_1+14&=8 \\ a_1&=-6\end{align*}$
::a1+(3-1)(7)=8a1+2(7)=8a1+14=8a1+6

Now that we have the first term and the common difference we can follow the same process used in the previous example to complete the problem.
::现在我们有了第一个术语和共同的区别,我们可以遵循上一个例子中用来解决问题的同样进程。

$\begin{aligned} a_{n} & = - 6 + (n - 1) (7) \\ = - 6 + 7 n - 7, thus a_{n} = 7 n - 13. \\ = 7 n - 13 \end{aligned}$
$\begin{align*}a_n&=-6+(n-1)(7) \\ &=-6+7n-7 \quad , \text{thus} \ a_n=7n-13. \\ &=7n-13\end{align*}$
::an6+(n- 1)(7) 6+7n-7n-7thus an=7n- 13.=7n-13

Now we can find the $\begin{align*}100^{th}\end{align*}$ term: $\begin{align*}a_{100}=7(100)-13=687\end{align*}$ .
::现在我们可以找到第100个术语:a100=7(100)-13=687。

Examples
::实例

Example 1
::例1

Earlier, you were asked to find the $\begin{align*}n^{th}\end{align*}$ term rule and the $\begin{align*}10^{th}\end{align*}$ term for the sequence represented by Halley's Comet, which appears in the sky once approximately every 76 years and first appeared in 1531.
::早些时候,有人要求你为Halley的彗星代表的序列找到 nth 术语规则和第十个术语规则, 它大约每隔76年出现在天空一次, 第一次出现在1531年。

From the information given, we can conclude that $\begin{align*}a_1=1531\end{align*}$ and $\begin{align*}d=76\end{align*}$ .
::根据所提供的资料,我们可以得出a1=1531和d=76的结论。

We now have what we need to plug into the rule:
::我们现在有我们需要的插入规则:

$\begin{aligned} a_{n} & = 1531 + (n - 1) (76) \\ = 1531 + 76 n - 76, thus the n^{t h} term rule is a_{n} = 76 n + 1455 \end{aligned}$
$\begin{align*}a_n&=1531+(n-1)(76) \\ &= 1531+76n-76 \quad , \text{thus the} \ n^{th} \ \text{term rule is} \ a_n=76n+1455\end{align*}$
::an=1531+(n-1)(76)=1531+76n-76,nth 术语规则为 an=76n+1455

Now to find the $\begin{align*}10^{th}\end{align*}$ term we can use our rule and replace $\begin{align*}n\end{align*}$ with 10: $\begin{align*}a_{10}=76(10) + 1455=760 + 1455 =2215\end{align*}$ .
::现在找到第10个术语, 我们可以使用我们的规则, 用 10: a10=76(10)+1455=760+1455=2215来替换 n。

Example 2
::例2

Find the common difference and the $\begin{align*}n^{th}\end{align*}$ term rule for the sequence: $\begin{align*}5, -3, -11,\ldots\end{align*}$
::找出共同的区别和Nth术语规则的顺序: 5 -3, -11,...

The common difference is $\begin{align*}-3-5=-8\end{align*}$ . Now $\begin{align*}a_n=5+(n-1)(-8)=5-8n+8=-8n+13\end{align*}$ .
::共同的差别是-3-58. 现在一个=5+(n-1)(-8)=5-8n+8+88n+13。

Example 3
::例3

Write the $\begin{align*}n^{th}\end{align*}$ term rule and find the $\begin{align*}45^{th}\end{align*}$ term for the arithmetic sequence with $\begin{align*}a_{10}=1\end{align*}$ and $\begin{align*}d=-6\end{align*}$ .
::以 a10=1 和 d6 来写入 nth 术语规则, 并找到算术序列的第 45 个术语。

To find the first term:
::为了找到第一个学期:

$\begin{aligned} a_{1} + (10 - 1) (- 6) & = 1 \\ a_{1} - 54 & = 1 \\ a_{1} & = 55 \end{aligned}$
$\begin{align*}a_1+(10-1)(-6)&=1 \\ a_1-54&=1 \\ a_1&=55\end{align*}$
::a1+(10-1)(-6)=1a1-54=1a1=1a1=55

Find the $\begin{align*}n^{th}\end{align*}$ term rule: $\begin{align*}a_n=55+(n-1)(-6)=55-6n+6=-6n+61\end{align*}$ .
::查找 nth 术语规则: an=55+(n- 1)(-6)=55- 6n+66n+61。

Finally, the $\begin{align*}45^{th}\end{align*}$ term: $\begin{align*}a_{45}=-6(45)+61=-209\end{align*}$ .
::最后,第45届:a456(45)+61209。

Example 4
::例4

Find the $\begin{align*}62^{nd}\end{align*}$ term for the arithmetic sequence with $\begin{align*}a_1=-7\end{align*}$ and $\begin{align*}d=\frac{3}{2}\end{align*}$ .
::以 a1++7 和 d=32 查找算术序列的第62 术语。

This time we will not simplify the $\begin{align*}n^{th}\end{align*}$ term rule, we will just use the formula to find the $\begin{align*}62^{rd}\end{align*}$ term: $\begin{align*}a_{62}=-7+(62-1) \left(\frac{3}{2}\right)=-7+61 \left(\frac{3}{2}\right)=- \frac{14}{2}+ \frac{183}{2}= \frac{169}{2}\end{align*}$ .
::这一次我们不会简化Nth术语规则, 我们只会使用公式来找到第62个术语: a627+(62-1)(32)+7+61(32)142+1832=1692。

Review
::回顾

Identify which of the following sequences is arithmetic. If the sequence is arithmetic find the $\begin{align*}n^{th}\end{align*}$ term rule.
::标明以下哪个序列是算术。如果序列是算术, 则查找 nth 术语规则。

$\begin{align*}2, 3, 4, 5, \ldots\end{align*}$

$\begin{align*}6, 2, -1, -3, \ldots\end{align*}$

$\begin{align*}5, 0, -5, -10, \ldots\end{align*}$

$\begin{align*}1, 2, 4, 8, \ldots\end{align*}$

$\begin{align*}0, 3, 6, 9, \ldots\end{align*}$

$\begin{align*}13, 12, 11, 10, \ldots\end{align*}$

$\begin{align*}4, -3, 2, -1, \ldots\end{align*}$

$\begin{align*}a, a+2, a+4, a+6, \ldots\end{align*}$
::a, a+2, a+4, a+6,...

Write the $\begin{align*}n^{th}\end{align*}$ term rule for each arithmetic sequence with the given term and common difference.
::以给定的术语和常见差数为每个算术序列写入 nth 术语规则。

$\begin{align*}a_1=15\end{align*}$ and $\begin{align*}d=-8\end{align*}$
::a1=15和d8

$\begin{align*}a_1=-10\end{align*}$ and $\begin{align*}d= \frac{1}{2}\end{align*}$
::a110和d=12

$\begin{align*}a_3=24\end{align*}$ and $\begin{align*}d=-2\end{align*}$
::a3=24和d2

$\begin{align*}a_5=-3\end{align*}$ and $\begin{align*}d=3\end{align*}$
::a53和d=3

$\begin{align*}a_{10}=-15\end{align*}$ and $\begin{align*}d=-11\end{align*}$
::a1015和d11

$\begin{align*}a_7=32\end{align*}$ and $\begin{align*}d=7\end{align*}$
::a7=32和d=7

$\begin{align*}a_{n-2}=3n+2\end{align*}$ , find $\begin{align*}a_n\end{align*}$
::an-2=3n+2, 找到 a

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

Section outline

General

具有共同差异和任期的理论性序列和查找第n期

Arithmetic Sequence ::亚学序列

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Review ::回顾

Review (Answers) ::回顾(答复)

Arithmetic Sequence
::亚学序列

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Review
::回顾

Review (Answers)
::回顾(答复)